AQA Qualifications
GCSE
MATHEMATICS (LINEAR)
4365/1H
Mark scheme
4365
June 2014
Version 1.0 Final
Mark schemes are prepared by the Lead Assessment Writer and considered, together with the
relevant questions, by a panel of subject teachers. This mark scheme includes any amendments
made at the standardisation events which all associates participate in and is the scheme which was
used by them in this examination. The standardisation process ensures that the mark scheme covers
the students’ responses to questions and that every associate understands and applies it in the same
correct way. As preparation for standardisation each associate analyses a number of students’
scripts: alternative answers not already covered by the mark scheme are discussed and legislated for.
If, after the standardisation process, associates encounter unusual answers which have not been
raised they are required to refer these to the Lead Assessment Writer.
It must be stressed that a mark scheme is a working document, in many cases further developed and
expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark
schemes on the basis of one year’s document should be avoided; whilst the guiding principles of
assessment remain constant, details will change, depending on the content of a particular
examination paper.
Further copies of this Mark Scheme are available from aqa.org.uk
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MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Glossary for Mark Schemes
GCSE examinations are marked in such a way as to award positive achievement wherever
possible. Thus, for GCSE Mathematics papers, marks are awarded under various categories.
M
Method marks are awarded for a correct method which could lead
to a correct answer.
A
Accuracy marks are awarded when following on from a correct
method. It is not necessary to always see the method. This can be
implied.
B
Marks awarded independent of method.
Q
Marks awarded for quality of written communication.
M dep
A method mark dependent on a previous method mark being
awarded.
B dep
A mark that can only be awarded if a previous independent mark
has been awarded.
ft
Follow through marks. Marks awarded for correct working
following a mistake in an earlier step.
SC
Special case. Marks awarded for a common misinterpretation
which has some mathematical worth.
oe
Or equivalent. Accept answers that are equivalent.
e.g. accept 0.5 as well as
1
2
[a, b]
Accept values between a and b inclusive.
[a, b)
Accept values a ≤ value < b
25.3 …
Allow answers which begin 25.3 e.g. 25.3, 25.31, 25.378.
Use of brackets
It is not necessary to see the bracketed work to award the marks.
3 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Examiners should consistently apply the following principles
Diagrams
Diagrams that have working on them should be treated like normal responses. If a diagram has been
written on but the correct response is within the answer space, the work within the answer space should be
marked. Working on diagrams that contradicts work within the answer space is not to be considered as
choice but as working, and is not, therefore, penalised.
Responses which appear to come from incorrect methods
Whenever there is doubt as to whether a candidate has used an incorrect method to obtain an answer, as a
general principle, the benefit of doubt must be given to the candidate. In cases where there is no doubt that
the answer has come from incorrect working then the candidate should be penalised.
Questions which ask candidates to show working
Instructions on marking will be given but usually marks are not awarded to candidates who show no working.
Questions which do not ask candidates to show working
As a general principle, a correct response is awarded full marks.
Misread or miscopy
Candidates often copy values from a question incorrectly. If the examiner thinks that the candidate has
made a genuine misread, then only the accuracy marks (A or B marks), up to a maximum of 2 marks are
penalised. The method marks can still be awarded.
Further work
Once the correct answer has been seen, further working may be ignored unless it goes on to contradict the
correct answer.
Choice
When a choice of answers and/or methods is given, mark each attempt. If both methods are valid then
M marks can be awarded but any incorrect answer or method would result in marks being lost.
Work not replaced
Erased or crossed out work that is still legible should be marked.
Work replaced
Erased or crossed out work that has been replaced is not awarded marks.
Premature approximation
Rounding off too early can lead to inaccuracy in the final answer. This should be penalised by 1 mark
unless instructed otherwise.
4 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Paper 1 Higher Tier
Answer
Q
Mark
1(a)
Expression
B1
1(b)
Formula or equation
B1
1(c)
Identity
B1
Comments
5 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Alternative Method 1
(CD = £) 45 – 35 or 10
or 2d + c = 35 and 2d + 2c = 45
M1
(35 – their 10) ÷ 2
or (45 – 2 × their 10) ÷ 2
or 22.5 – their 10 or 12.5(0)
M1
Condone missing brackets
or a pair of values that satisfy one of
the statements
3  their 10 + their 12.5(0)
M1dep
42.5(0) or 7.5(0) remaining
A1
Dep on second M
Strand (iii)
Correct conclusion based on their
total
ft correct conclusion based on their total if
two Ms awarded.
Q1ft
2
NB the difference between the cost of 3 CDs
and 50 may be calculated and compared to
the cost of a DVD to reach a conclusion. eg
50 – 3  10 = 20 > 12.5 so Yes is full marks.
Alternative Method 2 (Trial and Improvement)
Chooses a value for CD and DVD and
tests in both statements
M1
Chooses a new value for CD or DVD
or both and tests in both statements
M1dep
Finds a pair of values for CD and
DVD that the student thinks works in
both statements and calculates
3  their CD + their DVD
M1dep
42.5 (0)
Correct conclusion based on their
total
A1
Strand (iii)
Q1ft
ft correct conclusion based on their total if
three Ms awarded.
6 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
3(a)
Four different numbers in any order
with median 5 and range 7
B2
B1
Four numbers in any order with
median 5 and range 7 with repeats
eg 4, 4, 6, 11
eg 1, 4, 6, 8
3, 3, 7, 10
9, 6, 4, 2
1, 5, 5, 8
3, 10, 6, 4
5, 5, 4, 11
1, 3, 7, 8
B1
2, 3, 7, 9
Four different numbers in any order
with median 5 or range 7
0, 4, 6, 7
1.5, 4, 6, 8.5
–1, 4.5, 5.5, 6
7  6 or 42 or
3(b)
8  9 or 72 or
9  4 or 36 or
M1
10  1
At least one product shown or one correct
value (not 10)
or 160
Must have the sum of 4 products divided by
20.
(their 42 + their 72 + their 36 + their
10) ÷ 20
M1 dep
Condone missing brackets
(7  6 + 8  9 + 9  4 + 10 ( 1)) ÷ 20
is M2
8
A1
7 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Answer
Q
4(a)
4(b)
Mark
5×8
M1
40
A1
cm2
B1
Any quadrilateral that has neither
line nor rotational symmetry. Ie
Comments
oe ½ (8 + 8)  5
Rotations, translations and reflections of
these.
B1
Must use dots as vertices.
Condone internal lines if a clear quadrilateral
is outlined.
Lines do not need to be ruled.
5(a)
B1
for answer of 2 or 3 or 2  3
B1
for 24 = {2, 2, 2, 3} or 2223
or 42 = {2, 3, 7} or 2  3  7
or
6
B2
one pair of factors of 24 (not 1  24)
eg 2 × 12, 3 and 8, 24 ÷ 4 = 6 (oe)
and one pair of factors of 42
(not 1  42).
eg 2 × 21, 3/14, (6, 7) (oe)
or (24 =) {1, 2, 3, 4, 6, 8, 12, 24}
or (42 =) {1, 2, 3, 6, 7, 14, 21, 42}
8 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
5(b)
48 as a correct product (except 1 
48)
eg
M1
Ignore incorrect products if at least one
correct product seen.
2  24 or 3  16 or 6  8 or 2  3 
8 or (2, 24) or (1, 2, 3, 8) etc
6
2  2  2  2  3 or 24 × 3(1)
or 23  2  3(1) or 22  22  3(1)
A1
D, B, C
B2
π  5 or π  10 ÷ 2 or 2  π  5 ÷ 2
B1 for 2 correct
Accept numerical values eg 31 ÷ 2
M1
Allow π = 3
15.5
A1
Accept 15.7 or 15.71
25.5
A1ft
or 5π
7
oe eg 48 ÷ 2 = 24 or branches on a prime
factor tree showing at least one product or
factor ladder showing a correct division.
Answer
Q
ft 10 + their 15.5 Accept 25.7 or 25.71
SC2 5π + 10
Mark
6
B1
–6
B1
8(a)
Comments
Allow embedded answers ie
6  6 = 36, 62 = 36, -62 = 36, –6  –6 = 36
B2 for ±6
9 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Alternative method 1
Ignore any denominators. This is the
numerator of the left hand side.
8(b)
If expanded straight away allow one sign or
numerical error
2(y + 1) + 3(y – 2)
or 2y + 2 + 3y – 6
M1
Invisible brackets must be recovered for M1
eg 2  y + 1 + 3  y – 2 = 5y – 1 is M0
2(y + 1) + 3(y – 2) = 12
or 2y + 2 + 3y – 6 = 12 is M2
5y – 4
A1
Their 5y – 4 = 12
M1
3.2
A1ft
Could be implied by rearrangement
eg 5y = 6 from 2y + 2 + 3y – 6 = 2
oe eg 5y + 2 = 18
oe eg
ft on both Ms awarded and at
most 1 error.
Alternative method 2
M1
Allow one sign or numerical error
A1
M1
3.2
A1ft
This is for rearranging their LHS = 2 with
variable on one side and numbers on the
other with terms simplified.
oe eg
ft on both Ms awarded and at
most 1 error.
10 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
9(a)
a=6
B1
Allow 6x
SC1 if values reversed.
b = 100
B1
y = 6x + 100 seen in script with no
contradictory answers for a and b given
allow B2
Alternative Method 1
9(b)
y = their 6  80 + their 100.
Substitution of 80 into their formula
M1
580
A1ft
ft their formula
400 + (280 – 100)
M1
Or use of values from graph
580
A1
One comparison on means but must
clarify what this implies
B1
Their 6 must have a value, ie not 0.
Alternative Method 2
10
One comparison on interquartile
ranges but must clarify what this
implies
B1
eg mens mean is lower so they are faster
(on average)
Women are slower on average
Women are more consistent as their IQR is
smaller
Men’s times more varied as IQR bigger
11 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
11
Alternative Method 1
(P:) (D =) 90T
or (M:) (D =) 70(T + 1)
M1
oe
90T = 70(T + 1)
M1dep
Condone missing bracket, ie 90T = 70T + 1
but no further marks unless bracket
recovered,
90T – 70T = 70
M1dep
oe NB 70 ÷ 20 is M3
3.5
A1
oe 3.30 is M3, A0
Alternative Method 2
Chooses a value for distance travelled
and correctly works out time taken at
90kph and time taken at 70kph
Lists distance travelled for Paul and Mary (for
at least 2 hours)
M1
Eg 90, 180, 270, 360, …
70, 140, 210, 280, 350, ….
Subtracts their values or repeats
above with a different value
M1dep
Trying a new value implies that the
difference between previously calculated
times was not 1.
Chooses a different value for distance
travels and correctly works out time
taken at 90kph and time taken at
70kph, but the difference in times
must be closer to 1 hour than the
previous choice.
M1dep
oe
3.5
A1
oe 3.30 is M3, A0 SC2 315 km
Alternative Method 3
(P:) (D =) 90(t – 1)
or (M:) (D =) 70t
M1
NB this scheme is for working out the time
that Mary takes. It can be ‘recovered’ for full
marks but if it ends at 4.5 then 2 marks
maximum.
oe
20t = 90, and t = 4.5
M1dep
NB 90 ÷ 20 = 4.5 is M2
Their 4.5 – 1
M1dep
oe
3.5
A1
oe 3.30 is M3, A0
12 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Answer
Q
B
Mark
A, B
Comments
Mark the diagram first and only look in
working space if blanks in diagram.
0.12
0.4
A
0.3
0.6
Not B A, Not B
B
12
0.7
0.18
Not A, B 0.28
0.4
B
0.6
Not B Not A, Not B
0.42
P(B) = 0.4
B1
P(Not A) = 0.7 and 1 – their 0.4, and
same probabilities on both second
branches
B1ft
Any 1st event and 2nd event
probability multiplied together
M1
Follow through their values even if 0.7
wrong, but probabilities must be 0 < p < 1
Full correct final probabilities
A1ft
ft their probabilities
13 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Alternative Method 1
x + y = 15 and x – y = 8
M1
oe
Attempt to solve by eliminating x or y
M1
Eg 2x = 23
or 2y = 7
x = 11.5 or y = 3.5
A1
A1 ft
ft on one error but only for one variable.
eg if x calculated wrongly but then
substituted and y calculated correctly allow
A0 for x but A1 ft for y
y = 3.5 and x = 11.5
13
SC2 correct answers but no working or by
T&I
Alternative Method 2
13
a and 15 – a
M1
15 – a – a = 8
M1
a = 3.5
A1
b = 11.5
A1 ft
SC2 correct answers but no working or by
T&I
Alternative Method 3
b and b + 8
M1
b and b – 8
oe b + b + 8 = 15
M1
oe b + b – 8 = 15
b = 3.5
A1
b = 11.5
a = 11.5
A1 ft
a = 3.5
SC2 correct answers but no working or by
T&I
14 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Answer
Q
3x(4x + 1) – 2(6x – 3)
Mark
M1
14(a)
Comments
if expanded straight away allow one sign or
arithmetic error eg 12x2 + 3x – 12x – 6
(Must have an x2 term, 2 ‘x‘ terms and a
constant term)
Condone missing brackets eg
3x  4x + 1 – 2  6x – 3
12x2 + 3x – 12x + 6
A1
12x2 – 9x + 6 = 6
2
or 12x – 9x = 0
oe
M1
or 12x2 = 9x
If their equation in (a) is 12x2 – 9x – 6 leading
to 12x2 – 9x – 12 = 0 award M1
use of formula or completing the square
oe
14(b)
x(12x – 9) or 3x(4x – 3) or x(4x – 3)
or 12x = 9
M1dep
or (x –
)2 =
oe
from equation above
oe
3
x=
4
A1
If x = 0 given do not award A1
15 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
15
AD = AE (10 (cm) or sides of a
square) or sides marked as 10 on
diagram
B1
Must give a reason or mark sides as 10 on
diagram
AB = AG (10 (cm) or sides of a
square) or sides marked as 10 on
diagram
B1
Must give a reason or sides as 10 on
diagram
Angle DAG = angle EAB (135 or 90 +
45)
B1
Must state 135 or 90 + 45 or 135 shown for
both angles on diagram
Q1
Q0 for congruent without SAS, AAS etc or
the appropriate reason for their proof stated
in words (strand (ii))
Congruent due to SAS (could be
expressed in words eg two sides and
angle between them the same)
or congruent due to ASA or AAS or
SAA with 22.5 shown or stated (after
135 seen) as one of the other angles.
(could be in words eg two angles and
the side between them, or two angles
and a side)
Sight of x2, –xy, +xy and –y2 plus
some indication that xy terms cancel.
16(a)
Eg x2 – xy + xy – y2
B1
Minimum would be
x2 – xy + xy – y2 = x2 – y2
16 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Alternative method 1
16(b)
3 +1
1
 5√2  (√3 – 1) 
2
2 2
B1
Correct substitution into ½ absinC
( 3 – 1)( 3 + 1) = 3 – 1 (= 2)
B1
This must be evaluated at some stage
Q1
Must show or state cancelling (strand(ii)) for
justifying a result.
Clear indication that the expression
cancels down to a fraction equivalent
5
to
2
Cancelling can be done at any stage
Alternative method 2
Height = (√3 – 1) 
3 +1
2 2
=
B1
that ( 3 – 1)( 3 + 1) = 3 – 1 (= 2)
16(b)
1
 5√2 their
2
Clear indication that the expression
cancels down to a fraction equivalent
5
to
2
Must get this correct to show explicitly or
implicitly (eg could rationalise denominator)
B1ft
Q1
Must show or state cancelling (strand(ii)) for
justifying a result.
Cancelling can be done at any stage
17 of 18
MARK SCHEME – GCSE Mathematics (Linear) – 4365/1H – June 2014
Answer
Q
Mark
Comments
Alternative Method 1
Volume original =
1
 10  10  30
3
10  10 could be 102
M1
Accept 0.33 or better for
(= 1000)
Volume removed =
1
 5  5  15
3
5  5 could be 52
M1
1
 (10  10  30 – 5  5  15) is M2
3
A1
Correct answer only
(= 125)
875
17
Alternative Method 2
Volume original =
1
 10  10  30
3
10  10 could be 102
M1
Accept 0.33 or better for
(= 1000)
(Linear scale factor
1
so) volume
2
1
7
scale factor
or
8
8
18(a)
1
3
1
3
M1
875
A1
Correct answer only
Substitution of x = 0 into equation
M1
–
1
A1
1
 1  – 15
15
1
(2  their midpoint + 1)(2  their
15
midpoint – 15)
B1 for –
18(b)
–
1
(2  3.5 + 1)(2  3.5 – 15)
15
B2
or substituting into their expanded
expression.
B1 for graph intersects at x = –0.5 or
midpoint = 3.5
18 of 18